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  • Two equal drops of water are falling through air with a steady velocity <span style="font-family:"Cambria Math",se

Two equal drops of water are falling through air with a steady velocity v. If the drops coalesce, then new velocity will be

Option: 1

2v


Option: 2

\sqrt{2} v


Option: 3

2^{2 / 3} \cdot v


Option: 4

\frac{v}{\sqrt{2}}


Answers (1)

best_answer

Let r  be the radius of smaller drop and R  the radius of bigger drop. By equating the volumes, we have.

2\left(\frac{4}{3} \pi r^3\right)=\frac{4}{3} \pi R^3

R=\text { (2) } ^{1/3}\cdot r

Now, terminal velocity α (radius )2

v^{\prime}=2^{2 / 3} \cdot v

Posted by

rishi.raj

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